HILBERT-PACHPATTE TYPE INEQUALITIES FROM
BONSALL’S FORM OF HILBERT’S INEQUALITY
JOSIP PEČARIĆ
IVAN PERIĆ
Faculty of Textile Technology
University of Zagreb
Pierottijeva 6, 10000 Zagreb, Croatia
Faculty of Food Technology and Biotechnology
University of Zagreb
Pierottijeva 6, 10000 Zagreb, Croatia
EMail: pecaric@hazu.hr
EMail: iperic@pbf.hr
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
PREDRAG VUKOVIĆ
Faculty of Teacher Education
University of Zagreb, Ante Starčevića 55
40000 Čakovec, Croatia
Contents
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EMail: predrag.vukovic@vus-ck.hr
J
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Received:
19 June, 2007
Page 1 of 27
Accepted:
09 October, 2007
Communicated by:
B. Yang
2000 AMS Sub. Class.:
26D15.
Key words:
Inequalities, Hilbert’s inequality, sequences and functions, homogeneous kernels,
conjugate and non-conjugate exponents, the Beta function.
Abstract:
The main objective of this paper is to deduce Hilbert-Pachpatte type inequalities using Bonsall’s form of Hilbert’s and Hardy-Hilbert’s inequalities, both in
discrete and continuous case.
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Contents
1
Introduction
3
2
Integral Case
8
3
Discrete Case
17
4
Non-conjugate Exponents
23
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
Contents
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1.
Introduction
An interesting feature of one of the forms of Hilbert-Pachpatte type inequalities, is
that it controls the size (in the sense of Lp or lp spaces ) of the modified Hilbert
transform of a function or of a series with the size of its derivate or its backward
differences, respectively. We start with the following results of Zhongxue Lü from
[9], for both continuous and discrete cases. For a sequence a : N0 → R, the sequence
∇a : N → R is defined by ∇a(n) = a(n)−a(n−1). For a function u : (0, ∞) → R,
u0 denotes the usual derivative of u.
Theorem A. Let p > 1, p1 + 1q = 1, s > 2−min{p, q}, and f (x), g(y) be real-valued
continuous functions defined on [0, ∞), respectively, and let f (0) = g(0) = 0, and
Z ∞Z x
Z ∞Z y
1−s 0
p
0<
x |f (τ )| dτ dx < ∞, 0 <
y 1−s |g 0 (δ)|q dδdy < ∞,
0
0
0
(1.1)
0
∞
∞
|f (x)||g(y)|
dxdy
(qxp−1 + py q−1 )(x + y)s
0
q+s−2 p+s−2
Z ∞ Z x
p1
B
, p
q
·
x1−s |f 0 (τ )|p dτ dx
≤
pq
0
0
Z ∞ Z y
1q
1−s 0
q
×
y |g (δ)| dδdy .
Z
0
Theorem B. Let p > 1,
1
p
+
1
q
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
Contents
0
then
Z
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
0
= 1, s > 2 − min{p, q}, and {a(m)} and {b(n)} be
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two sequences of real numbers where m, n ∈ N0 , and a(0) = b(0) = 0, and
0<
∞ X
m
X
m=1 τ =1
m
1−s
p
|∇a(τ )| < ∞,
0<
∞ X
n
X
n1−s |∇b(δ)|q < ∞,
n=1 δ=1
then
(1.2)
∞ X
∞
X
m=1 n=1
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
|am ||bn |
p−1
(qm
+ pnq−1 )(m + n)s
! p1
p+s−2
∞ X
m
B q+s−2
,
X
q
p
≤
·
m1−s |∇a(τ )|p
pq
m=1 τ =1
×
∞ X
n
X
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
! 1q
n1−s |∇b(δ)|q
n=1 δ=1
Note that the condition s > 2 − min{p, q} from Theorem B is not sufficient.
Namely, the author of the proof of Theorem B used the following result
∞
m 2−s
X
1
q+s−2 p+s−2
q
(1.3)
,
m1−s ,
<B
s
(m
+
n)
n
q
p
n=1
for m ∈ {1, 2, . . .} and s > 2 − min{p, q}. For p = q = 2, s = 18 and m = 1,
the left-hand side of (1.3) is greater than the right-hand side of (1.3). Therefore, we
refer to a paper of Krnić and Pečarić, [4], where the next inequality is given:
(1.4)
∞
X
n=1
1
m α1
< m1−s+α1 −α2 B(1 − α2 , s + α2 − 1),
(m + n)s nα2
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.
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where 0 < s ≤ 14, 1 − s < α2 < 1 for s ≤ 2 and −1 ≤ α2 < 1 for s > 2. By using
this result, here we shall obtain a generalization of Theorem B but with the condition
2 − min{p, q} < s ≤ 2 + min{p, q}. Also, the following result is given in [9]:
∞
m 2−s
X
1
q+s−2 p+s−2
1
q
(1.5)
< B
,
m1−s ,
s + ns
m
n
s
sq
sp
n=1
for m ∈ {1, 2, . . .} and s > 2 − min{p, q}. Similarly as before, for p = q = 2, s = 6
and m = 1, the left-hand side of (1.5) is greater than the right-hand side of (1.5).
The case of nontrivial weights is essential in Theorem A and Theorem B, since for
s = 1 only the trivial functions and sequences satisfy the assumptions.
In 1951, Bonsall established the following conditions for non-conjugate exponents (see [1]). Let p and q be real parameters, such that
(1.6)
p > 1, q > 1,
(1.7)
λ=
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
Contents
1 1
+ ≥ 1,
p q
and let p0 and q 0 respectively be their conjugate exponents, that is,
1
+ q10 = 1. Further, define
q
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
1
p
+
1
p0
= 1 and
1
1
+ 0
0
p
q
and note that 0 < λ ≤ 1 for all p and q as in (1.6). In particular, λ = 1 holds if and
only if q = p0 , that is, only when p and q are mutually conjugate. Otherwise, we have
0 < λ < 1, and in such cases p and q will be referred to as non-conjugate exponents.
Also, in this paper we shall obtain some generalizations of (1.1). It will be done in
simplier way than in [9]. Our results will be based on the following results of Pečarić
et al., [2], for the non-conjugate and conjugate exponents.
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Theorem C. Let p, p0 , q, q 0 and λ be as in (1.6) and (1.7). If K, ϕ, ψ, f and g
are non-negative measurable functions, then the following inequalities hold and are
equivalent
Z
p1 Z
1q
Z
λ
q
p
(1.8)
K (x, y)f (x)g(y)dxdy ≤
(ϕF f ) (x)dx
(ψGg) (y)dy
Ω2
Ω
Ω
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and
Z (1.9)
Ω
1
ψG(y)
Z
K λ (x, y)f (x)dx
! 10
q 0
q
Z
(ϕF f )p (x)dx
≤
dy
Ω
and Predrag Vuković
p1
,
Ω
where the functions F, G are defined by
Z
10
q
K(x, y)
and
F (x) =
dy
q0
Ω ψ (y)
Title Page
Z
K(x, y)
dx
ϕp0 (x)
G(y) =
Ω
Contents
10
p
.
The next inequalities from [5] can be seen as a special case of (1.8) and (1.9)
respectively for the conjugate exponents:
Z
(1.10)
K(x, y)f (x)g(y)dxdy
Ω2
ϕp (x)F (x)f p (x)dx
≤
p1 Z
Ω
ψ q (y)G(y)g q (y)dy
1q
Ω
and
G
(1.11)
Ω
1−p
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Z
Z
vol. 9, iss. 1, art. 9, 2008
−p
p
Z
(y)ψ (y)
K(x, y)f (x)dx
Ω
Z
dy ≤
Ω
ϕp (x)F (x)f p (x)dx,
Close
where
Z
F (x) =
(1.12)
Ω
K(x, y)
dy
ψ p (y)
Z
and G(y) =
Ω
K(x, y)
dx.
ϕq (x)
In particular, inequalities (1.10) and (1.11) are equivalent.
On the other hand, here we also refer to a paper of Brnetić et al., [8], where a
general Hilbert-type inequality was obtained for
Pnn ≥1 2 conjugate exponents, that
is, real parameters p1 , . . . , pn > 1, such that i=1 pi = 1. Namely, we let K :
n
Ω
Qn → R and φij : Ω → R, i, j = 1, . . . , n, be non-negative measurable functions. If
i,j=1 φij (xj ) = 1, then the inequality
Z
K(x1 , . . . , xn )
(1.13)
Ωn
n
Y
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
fi (xi ) dx1 . . . dxn
Contents
i=1
≤
n Z
Y
i=1
Fi (xi )(φii fi )pi (xi ) dxi
p1
i
,
Ω
holds for all non-negative measurable functions f1 , . . . , fn : Ω → R, where
Z
n
Y
(1.14) Fi (xi ) =
K(x1 , . . . , xn )
φpiji (xj )dx1 . . . dxi−1 dxi+1 . . . dxn ,
Ωn−1
for i = 1, . . . , n.
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
j=1,j6=i
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2.
Integral Case
In this section we shall state our main results. We suppose that all integrals converge
and shall omit these types of conditions. Thus, we have the following.
Theorem 2.1. Let p1 + 1q = 1 with p > 1. If K(x, y), ϕ(x), ψ(y) are non-negative
functions and f (x), g(y) are absolutely continuous functions such that f (0) =
g(0) = 0, then the following inequalities hold
Z ∞Z ∞
K(x, y)|f (x)| |g(y)|
(2.1)
dxdy
qxp−1 + py q−1
0
0
Z ∞Z ∞
1 1
≤
K(x, y)|f (x)| |g(y)|d x p d y q
0
≤
1
pq
0
0
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
p1
x
p
0
p
ϕ (x)F (x)|f (τ )| dτ dx
∞Z
Z
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
Contents
0
∞
Z
Z
×
0
y
ψ q (y)G(y)|g 0 (δ)|q dδdy
1q
,
0
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and
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Z
(2.2)
0
∞
G1−p (y)ψ −p (y)
Z
0
∞
p
1
K(x, y)|f (x)|d x p
dy
Z Z
1 ∞ x p
≤ p
ϕ (x)F (x)|f 0 (τ )|p dτ dx,
p 0
0
where F (x) and G(y) are defined as in (1.12).
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Proof. By using Hölder’s inequality, (see also [9]), we have
(2.3)
1
q
|f (x)| |g(y)| ≤ x y
1
p
p1 Z
x
Z
0
p
1q
y
0
|f (τ )| dτ
q
|g (δ)| dδ
0
.
0
From (2.3) and using the elementary inequality
xy ≤
(2.4)
1 1
xp y q
+ , x ≥ 0, y ≥ 0,
+ = 1, p > 1,
p
q
p q
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
we observe that
(2.5)
vol. 9, iss. 1, art. 9, 2008
pq|f (x)| |g(y)|
|f (x)| |g(y)|
≤
≤
1
1
p−1
q−1
qx
+ py
xq y p
and therefore
Z ∞Z
(2.6) pq
0
Z
1
p
x
0
Z
p
y
0
q
1
q
|g (δ)| dδ
|f (τ )| dτ
0
0
Title Page
Contents
∞
K(x, y)|f (x)| |g(y)|
dxdy
qxp−1 + py q−1
0
Z ∞Z ∞
K(x, y)
≤
1
1 |f (x)||g(y)|dxdy
0
0
xq y p
Z x
p1 Z
Z ∞Z ∞
0
p
≤
K(x, y)
|f (τ )| dτ
0
0
0
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Page 9 of 27
y
|g 0 (δ)|q dδ
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1q
dxdy.
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Applying the substitutions
Z
x
0
p
|f (τ )| dτ
f1 (x) =
0
p1
Z
,
y
0
q
|g (δ)| dδ
g1 (y) =
0
1q
and (1.10), we obtain
Z ∞Z ∞
(2.7)
K(x, y)f1 (x)g1 (y)dxdy
0
0
Z
∞
≤
ϕ
p
(x)F (x)f1p (x)dx
p1 Z
ψ
0
Z
q
(y)G(y)g1q (y)dy
1q
0
∞
Z
x
=
0
∞
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
p1
p
0
p
ϕ (x)F (x)|f (τ )| dτ dx
and Predrag Vuković
0
Z
∞Z
y
ψ q (y)G(y)|g 0 (δ)|q dδdy
×
0
1q
vol. 9, iss. 1, art. 9, 2008
.
0
By using (2.6) and (2.7) we obtain (2.1). The second inequality (2.2) can be proved
by applying (1.11) and the inequality
|f (x)| ≤ x
1
q
Z
x
p1
|f (t)| dt .
0
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p
0
Page 10 of 27
Now we can apply our main result to non-negative homogeneous functions. Recall that for a homogeneous function of degree −s, s > 0, the equality K(tx, ty) =
t−s K(x, y) is satisfied. Further, we define
Z ∞
k(α) :=
K(1, u)u−α du
0
and suppose that k(α) < ∞ for 1 − s < α < 1. To prove first application of our
main results we need the following lemma.
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Lemma 2.2. If s > 0, 1 − s < α < 1 and K(x, y) is a non-negative homogeneous
function of degree −s, then
α
Z ∞
x
(2.8)
K(x, y)
dy = x1−s k(α).
y
0
Proof. By using the substitution u = xy and the fact that K(x, y) is homogeneous
function, the equation (2.8) follows easily.
Corollary 2.3. Let s > 0, p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely
continuous functions such that f (0) = g(0) = 0, and K(x, y) is a non-negative
symmetrical and homogeneous function of degree −s, then the following inequalities
hold
Z ∞Z ∞
K(x, y)|f (x)| |g(y)|
(2.9)
dxdy
qxp−1 + py q−1
0
0
Z ∞Z ∞
1 1
≤
K(x, y)|f (x)| |g(y)|d x p d y q
0
L
≤
pq
0
∞
Z
x
0
p1
x
Z
1−s+p(A1 −A2 )
0
p
|f (τ )| dτ dx
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
Contents
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Page 11 of 27
0
Z
∞
Z
×
y
y
0
1−s+q(A2 −A1 )
0
q
1q
|g (δ)| dδdy
0
and
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Z
(2.10)
0
∞
y (p−1)(s−1)+p(A1 −A2 )
p
∞
1
K(x, y)|f (x)|d(x p ) dy
0
p Z ∞ Z x
L
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
p
0
0
Z
1
1
where A1 ∈ ( 1−s
, 1q ), A2 ∈ ( 1−s
, p1 ) and L = k(pA2 ) p k(qA1 ) q .
q
p
Proof. Let F (x), G(y) be the functions defined as in (1.12). Setting ϕ(x) = xA1 and
ψ(y) = y A2 , by Lemma 2.2 we obtain
Z ∞Z x
(2.11)
ϕp (x)F (x)|f 0p dτ dx
0
0
pA2 !
Z ∞
Z ∞Z x
x
dy xp(A1 −A2 ) dτ dx
=
|f 0 (τ )|p
K(x, y)
y
0
0
0
Z ∞Z x
= k(pA2 )
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
0
and similarly
Z ∞Z
(2.12)
0
0
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
Contents
y
ψ q (y)G(y)|g 0 (δ)|q dδdy
0
Z
∞
Z
= k(qA1 )
0
y
y 1−s+q(A2 −A1 ) |g 0 (δ)|q dδdy.
0
From (2.1), (2.11) and (2.12), we get (2.9). Similarly, the inequality (2.10) follows
from (2.2).
We proceed with some special homogeneous functions. First, by putting K(x, y) =
in Corollary 2.3, we get the following.
1
(x+y)s
Corollary 2.4. Let s > 0, p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely
continuous functions such that f (0) = g(0) = 0, then the following inequalities
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hold
∞
Z
0
Z
0
∞
|f (x)| |g(y)|
dxdy
+ py q−1 )(x + y)s
Z ∞Z ∞
|f (x)| |g(y)| p1 1q ≤
d x d y
(x + y)s
0
0
Z ∞ Z x
p1
L1
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx
pq
0
0
Z ∞ Z y
1q
1−s+q(A2 −A1 ) 0
q
×
y
|g (δ)| dδdy
(qxp−1
0
and
Z
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
0
Title Page
∞
y
(p−1)(s−1)+p(A1 −A2 )
0
Z
0
where A1 ∈
( 1−s
, 1q ),
q
A2 ∈
∞
p
|f (x)| p1 d x
dy
(x + y)s
p Z ∞ Z x
L1
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
p
0
0
( 1−s
, p1 )
p
and
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1
1
L1 = [B(1 − pA2 , pA2 + s − 1)] p [B(1 − qA1 , qA1 + s − 1)] q .
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Remark 1. By putting A1 = A2 = 2−s
in Corollary 2.4, with the condition s >
pq
2 − min{p, q}, we obtain Theorem A from the introduction.
Close
ln
y
Since the function K(x, y) = y−xx is symmetrical and homogeneous of degree
−1, by using Corollary 2.3 we obtain:
Corollary 2.5. Let p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely continuous
functions such that f (0) = g(0) = 0, then the following inequalities hold
Z ∞Z ∞
ln xy |f (x)| |g(y)|
dxdy
p−1 + py q−1 )(y − x)
0
0 (qx
Z ∞Z ∞ y
ln x |f (x)| |g(y)| 1 1 ≤
d xp d y q
y−x
0
0
Z ∞ Z x
1q
p1 Z ∞ Z y
L2
q(A2 −A1 ) 0
q
p(A1 −A2 ) 0
p
|g (δ)| dδdy
|f (τ )| dτ dx
≤
y
x
pq
0
0
0
0
and
Z ∞
y
0
p(A1 −A2 )
Z
0
∞
|f (x)| ln xy 1 d xp
y−x
p
dy ≤
L2
p
p Z
∞
Z
p(A1 −A2 )
0
|f (τ )| dτ dx,
2
Similarly, for the symmetrical homogeneous function of degree −s, K(x, y) =
1
, we have:
max{x,y}s
1
q
Contents
0
L2 = π 2 (sin pA2 π)− p (sin qA1 π)− q .
1
p
vol. 9, iss. 1, art. 9, 2008
p
where A1 ∈ (0, 1q ), A2 ∈ (0, p1 ) and
2
and Predrag Vuković
Title Page
x
x
0
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
Corollary 2.6. Let s > 0, + = 1 with p > 1. If f (x), g(y) are absolutely
continuous functions such that f (0) = g(0) = 0, then the following inequalities
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hold
Z
0
∞
Z
0
∞
|f (x)| |g(y)|
dxdy
(qxp−1 + py q−1 ) max{x, y}s
Z ∞Z ∞
|f (x)| |g(y)| p1 1q ≤
d x d y
max{x, y}s
0
0
Z ∞ Z x
p1
L3
1−s+p(A1 −A2 ) 0
p
≤
x
|f (τ )| dτ dx
pq
0
0
Z ∞ Z y
1q
1−s+q(A2 −A1 ) 0
q
y
|g (δ)| dδdy
×
0
and
Z
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
0
Title Page
∞
Z
∞
p
1
|f (x)|
d
xp
dy
max{x, y}s
0
0
p Z ∞ Z x
L3
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
p
0
0
1
1
where A1 ∈ 1−s
, 1q , A2 ∈ 1−s
, p1 and L3 = k(pA2 ) p k(qA1 ) q , where k(α) =
q
p
s
.
(1−α)(s+α−1)
y (p−1)(s−1)+p(A1 −A2 )
At the end of this section we give a generalization of the inequality (2.1) from
Theorem 2.1. In the proof we used a general Hilbert-type inequality (1.13) of Brnetić
et al., [8].
P
Theorem 2.7. Let n ∈ N, n ≥ 2, ni=1 p1i = 1 with pi > 1, i = 1, . . . , n. Let qi , αi ,
Q
i = 1, . . . , n, are defined with q1i = 1 − p1i and αi = nj=1,j6=i pj . If K(x1 , . . . , xn ),
Q
φij (xj ), i, j = 1, . . . , n, are non-negative functions such that ni,j=1 φij (xj ) = 1,
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Page 15 of 27
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and fi (xi ), i = 1, . . . , n, are absolutely continuous functions such that fi (0) = 0,
i = 1, . . . , n, then the following inequality holds
Q
Z ∞
Z ∞
K(x1 , . . . , xn ) ni=1 |fi (xi )|
...
dx1 . . . dxn
Pn
pi −1
0
0
i=1 αi xi
1
1
Z ∞
Z ∞
n
Y
p1
≤
...
K(x1 , . . . , xn )
|fi (xi )|d x1 . . . d xnpn
0
0
i=1
1
p
n Z ∞ Z xi
Y
i
1
0
pi
pi
,
φii (xi )Fi (xi )|fi (τi )| dτi dxi
≤
p1 . . . pn i=1
0
0
where Fi (xi ) are defined as in (1.14) for i = 1, . . . , n.
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
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3.
Discrete Case
We also give results for the discrete case. For that, we apply the following result
from [5].
Theorem 3.1. If {a(m)} and {b(n)} are non-negative real sequences, K(x, y) is
non-negative homogeneous function of degree −s strictly decreasing in both parameters x and y, p1 + 1q = 1, p > 1, A, B, α, β > 0, then the following inequalities
hold and are equivalent
(3.1)
∞ X
∞
X
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
K(Amα , Bnβ )am bn
m=1 n=1
<N
∞
X
! p1
Title Page
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) apm
Contents
m=1
∞
X
·
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β) bqn
,
n=1
∞
X
∞
X
nβ(s−1)(p−1)+pβ(A1 −A2 )+β−1
n=1
!p
J
I
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K(Amα , Bnβ )am
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m=1
<N
p
∞
X
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) apm ,
m=1
where A1 ∈
(3.3)
II
Page 17 of 27
and
(3.2)
JJ
(max{ 1−s
, α−1
}, 1q ),
q
αq
1
1
N = α− q β − p A
A2 ∈ max
2−s
+A1 −A2 −1
p
B
n
1−s β−1
, βp
p
2−s
+A2 −A1 −1
q
o
,
1
p
1
and
1
k(pA2 ) p k(2 − s − qA1 ) q .
Close
Applying Theorem 3.1, we obtain the following.
Corollary 3.2. Let s > 0, p1 + 1q = 1 with p > 1. Let {a(m)} and {b(n)} be two
sequences of real numbers where m, n ∈ N0 , and a(0) = b(0) = 0. If K(x, y)
is a non-negative homogeneous function of degree −s strictly decreasing in both
parameters x and y, A, B, α, β > 0, then the following inequalities hold
∞ X
∞
X
K(Amα , Bnβ )|am | |bn |
m=1 n=1
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
qmp−1 + pnq−1
and Predrag Vuković
∞ ∞
1 X X K(Amα , Bnβ )|am | |bn |
≤
1
1
pq m=1 n=1
mq np
(3.4)
N
<
pq
·
∞ X
m
X
vol. 9, iss. 1, art. 9, 2008
! p1
m
α(1−s)+αp(A1 −A2 )+(p−1)(1−α)
Title Page
p
|∇a(τ )|
Contents
m=1 τ =1
n
∞ X
X
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β) |∇b(δ)|q
,
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n=1 δ=1
Page 18 of 27
and
(3.5)
∞
X
nβ(s−1)(p−1)+pβ(A1 −A2 )+β−1
n=1
∞
X
m=1
< Np
∞ X
m
X
K(Amα , Bnβ )
|am |
!p
1
n
o
Full Screen
mq
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) |∇a(τ )|p ,
m=1 τ =1
Go Back
n
o α−1
1
1−s β−1
where A1 ∈ max 1−s
,
,
,
A
∈
max
,
, p1 and the constant
2
q
αq
q
p
βp
N is defined as in (3.3).
Close
Proof. By using Hölder’s inequality, (see also [9]), we have
1
q
|am | |bn | ≤ m n
(3.6)
1
p
m
X
! p1
n
X
|∇a(τ )|p
τ =1
! 1q
|∇b(δ)|q
.
δ=1
From (2.4) and (3.6), we get
(3.7)
pq|am | |bn |
|am | |bn |
≤
≤
1
1
qmp−1 + pnq−1
mq np
m
X
! p1
|∇a(τ )|p
τ =1
n
X
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
! 1q
|∇b(δ)|q
,
and Predrag Vuković
δ=1
vol. 9, iss. 1, art. 9, 2008
and therefore
(3.8) pq
Title Page
∞ X
∞
X
K(Amα , Bnβ )|am | |bn |
qmp−1 + pnq−1
m=1 n=1
≤
∞ X
∞
X
K(Amα , Bnβ )|am | |bn |
1
m=1 n=1
≤
Contents
∞ X
∞
X
1
mq np
K(Amα , Bnβ )
m=1 n=1
m
X
τ =1
! p1
|∇a(τ )|p
n
X
! 1q
|∇b(δ)|q
(3.9)
m=1 n=1
K(Amα , Bnβ )e
amebn
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Page 19 of 27
.
δ=1
1
1
P
Pn
p p e
q q
Applying the substitutions e
am = ( m
τ =1 |∇a(τ )| ) , bn = (
δ=1 |∇b(δ)| ) and
(3.1), we have
∞ X
∞
X
JJ
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∞
X
<N
! p1
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α)e
apm
m=1
·
∞
X
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β)ebqn
,
n=1
=
∞ X
m
X
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
! p1
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) |∇a(τ )|p
and Predrag Vuković
m=1 τ =1
·
∞ X
n
X
vol. 9, iss. 1, art. 9, 2008
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β) |∇b(δ)|q
,
n=1 δ=1
Title Page
where A1 ∈ (max{ 1−s
, α−1
}, 1q ), A2 ∈ (max{ 1−s
, β−1
}, p1 ) and the constant N is
q
αq
p
βp
defined as in (3.3). Now, by applying (3.8) and (3.9) we obtain (3.4). The second
inequality (3.5) can be proved by using (3.2) and the inequality
! p1
m
X
1
|am | ≤ m q
|∇a(τ )|p
.
τ =1
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Remark 2. If the function K(x, y) from the previous corollary is symmetrical, then
1
2−s
k(2 − s − qA1 ) = k(qA1 ). So, if K(x, y) = (x+y)
s , then we can put A1 = A2 = pq ,
A = B = α = β = 1 in Corollary 3.2 and obtain Theorem B from the introduction
but with the condition 2 − min{p, q} < s < 2.
By using (1.4), see [4], we will obtain a larger interval for the parameter s. More
precisely, we have:
Close
Corollary 3.3. Let p1 + 1q = 1 with p > 1 and 2 − min{p, q} < s ≤ 2 + min{p, q}.
Let {a(m)} and {b(n)} be two sequences of real numbers where m, n ∈ N0 , and
a(0) = b(0) = 0. Then the following inequalities hold
(3.10)
∞ X
∞
X
m=1 n=1
(qmp−1
|am | |bn |
+ pnq−1 )(m + n)s
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
∞ ∞
1 XX
|am | |bn |
≤
1
pq m=1 n=1 m q n p1 (m + n)s
N1
<
pq
∞ X
m
X
and Predrag Vuković
! p1
m
1−s
p
∞ X
n
X
·
|∇a(τ )|
m=1 τ =1
! 1q
n
1−s
q
|∇b(δ)|
vol. 9, iss. 1, art. 9, 2008
,
n=1 δ=1
Title Page
and
∞
X
(3.11)
n(s−1)(p−1)
!p
|am |
1
q
m=1 m (m +
n=1
where N1 =
∞
X
n)s
∞ X
m
X
p
< N1
Contents
m1−s |∇a(τ )|p ,
m=1 τ =1
B( s+q−2
, s+p−2
).
q
p
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Page 21 of 27
Proof. As in the proof of Corollary 3.2, by using Hölder’s inequality we obtain
∞ X
∞
X
Go Back
|am | |bn |
p−1
(qm
+ pnq−1 )(m + n)s
m=1 n=1
∞ X
∞
X
1
≤
pq
Close
|am | |bn |
1
m=1 n=1
Full Screen
1
m q n p (m + n)s
∞ ∞
1 XX
1
≤
pq m=1 n=1 (m + n)s
m
X
τ =1
! p1
p
|∇a(τ )|
n
X
δ=1
! 1q
q
|∇b(δ)|
m 2−s
n 2−s
pq
pq
n
m
1
≤
pq
∞
∞ X
m
X
X
m=1 τ =1
·
n=1
∞ X
n
X
n=1 δ=1
m 2−s
1
q
(m + n)s n
∞
X
! p1
!
|∇a(τ )|p
n 2−s
1
p
(m + n)s m
m=1
! 1q
!
|∇b(δ)|q
.
Now, the inequality (3.10) follows from (1.4). Let us show that the inequality (3.11)
is valid. For this purpose we use the following inequality from [4]
!p
∞
∞
∞
X
X
X
a
m
(s−1)(p−1)
(3.12)
n
< L1
m1−s apm ,
s
(m
+
n)
n=1
m=1
m=1
where 2 − min{p, q} < s ≤ 2 + min{p, q} and L1 =
m
X
am =
B( s+p−2
, s+q−2
).
p
q
! p1
p
|∇a(τ )|
τ =1
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
Title Page
Setting
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Page 22 of 27
in (3.12) and using
|am | ≤ m
1
q
m
X
! p1
p
|∇a(τ )|
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,
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τ =1
the inequality (3.11) follows easily.
Close
4.
Non-conjugate Exponents
Let p, p0 , q, q 0 and λ be as in (1.6) and (1.7). To obtain an analogous result for
the case of non-conjugate exponents, we introduce real parameters r0 , r such that
p ≤ r0 ≤ q 0 and r10 + 1r = 1. For example, we can define r10 = q10 + 1−λ
or r0 = (2−λ)p.
2
It is easy to see that
0
1
r
r
1
1
0
0
0
0
0
(4.1)
x p y q ≤ 0 rx p + r y q ,
x ≥ 0, y ≥ 0,
rr
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
and
(4.2)
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
1
p0
|f (x)| |g(y)| ≤ x y
1
q0
Z
x
0
p
p1 Z
0
q
1q
|g (δ)| dδ
|f (τ )| dτ
0
y
,
Title Page
0
Contents
hold, where f (x), g(y) are absolutely continuous functions on (0, ∞).
Applying Theorem C, (4.1) and (4.2) in the same way as in the proof of Theorem
2.1, we obtain the following result for non-conjugate exponents.
0
0
0
Theorem 4.1. Let p, p , q, q and λ be as in (1.6) and (1.7). Let r , r be real
parameters such that p ≤ r0 ≤ q 0 and r10 + 1r = 1. If K(x, y), ϕ(x), ψ(y) are
non-negative functions and f (x), g(y) are absolutely continuous functions such that
f (0) = g(0) = 0, then the following inequalities hold
Z ∞Z ∞ λ
K (x, y)|f (x)| |g(y)|
(4.3)
dxdy
r0
r
0
0
rx p0 + r0 y q0
Z Z
1 1
pq ∞ ∞ λ
≤ 0
K (x, y)|f (x)| |g(y)|d x p d y q
rr 0
0
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1
≤ 0
rr
∞
Z
p1
x
Z
p
0
p
(ϕF ) (x)|f (τ )| dτ dx
0
0
Z
∞
Z
×
1q
y
0
q
q
(ψG) (y)|g (δ)| dδdy
0
,
0
and
Z
(4.4)
0
∞
1
ψG(y)
Z
∞
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
q0 ! q10
dy
K λ (x, y)|f (x)|d(x )
1
p
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
0
1
≤
p
Z
∞
Z
0
x
p1
(ϕF ) (x)|f (τ )| dτ dx ,
p
0
p
0
Title Page
Contents
where F (x) and G(y) are defined as in Theorem C.
Obviously, Theorem 4.1 is the generalization of Theorem 2.1. Namely, if λ = 1,
0
r = p and r = q, then the inequalities (4.3) and (4.4) become respectively the
inequalities (2.1) and (2.2). If K(x, y) is a non-negative symmetrical and homogeneous function of degree −s, s > 0, then we obtain:
Corollary 4.2. Let s > 0, p, p0 , q, q 0 and λ be as in (1.6) and (1.7). If f (x), g(y) are
absolutely continuous functions such that f (0) = g(0) = 0, and K(x, y) is a nonnegative symmetrical and homogeneous function of degree −s, then the following
inequalities hold
Z ∞Z ∞
K λ (x, y)|f (x)| |g(y)|
(4.5)
dxdy
(p−1)(2−λ) + py (q−1)(2−λ)
0
0 qx
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1
≤
2−λ
∞
Z
∞
Z
0
1 1
K λ (x, y)|f (x)| |g(y)|d x p d y q
0
M
≤
pq(2 − λ)
Z
·
∞
Z
x
Z
x
0
∞
p1
|f (τ )| dτ dx
0
p
0
y
Z
y
0
p
(1−s)+p(A1 −A2 )
q0
q
(1−s)+q(A2 −A1 )
p0
0
1q
q
|g (δ)| dδdy
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
0
and
and Predrag Vuković
Z
∞
y
(4.6)
q0
(s−1)+q 0 (A1 −A2 )
p0
Z
0
q 0
1
K λ (x, y)|f (x)|d x p
vol. 9, iss. 1, art. 9, 2008
! 10
q
dy
0
≤
where A1 ∈
∞
1−s 1
, p0
p0
, A2 ∈
M
p
Title Page
∞Z
Z
1−s 1
, q0
q0
x
p
x q0
0
(1−s)+p(A1 −A2 )
|f 0 (τ )|p dτ dx
p1
0
1
1
and M = k(p0 A1 ) p0 k(q 0 A2 ) q0 .
Proof. The proof follows directly from Theorem 4.1 setting r0 = (2 − λ)p, r =
(2 − λ)q, ϕ(x) = xA1 and ψ(y) = y A2 in the inequalities (4.3) and (4.4). Namely, if
F (x) and G(y) are the functions defined by
Z
F (x) =
0
∞
K(x, y)
dy
ψ q0 (y)
10
Z
q
and
G(y) =
0
∞
K(x, y)
dx
ϕp0 (x)
10
p
,
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then applying Lemma 2.2 we have
Z
p
pA1
(4.7)
(ϕF ) (x) = x
∞
K(x, y)y
−q 0 A2
p0
q
dy
0
= xpA1 −pA2
Z
0
=x
∞
q0 A2 ! qp0
x
K(x, y)
dy
y
p
(1−s)+p(A1 −A2 )
q0
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
p
q0
k (q 0 A2 ),
and Predrag Vuković
vol. 9, iss. 1, art. 9, 2008
and similarly
q
(4.8)
(ψG)q (y) = y p0
(1−s)+q(A2 −A1 )
q
k p0 (p0 A1 ).
Title Page
Now, by using (4.3), (4.7) and (4.8) we obtain (4.5).
The second inequality (4.6) follows directly from (4.4).
1
in Corollary 4.2 we obtain that the constant
(x+y)s
1
1
p0 A1 , p0 A1 + s − 1) p0 B(1 − q 0 A2 , q 0 A2 + s − 1) q0 .
Remark 3. Setting K(x, y) =
is equal to M = B(1 −
Contents
M
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References
[1] F.F. BONSALL, Inequalities with non-conjugate parameters, Quart. J. Math.
Oxford Ser. (2), 2 (1951), 135–150.
[2] I. BRNETIĆ, M. KRNIĆ AND J. PEČARIĆ, Multiple Hilbert’s and HardyHilbert’s integral inequality with non-conjugate parameters, Bull. Austral. Math.
Soc., 71 (2005), 447–457.
Hilbert-Pachpatte Type
Inequalities
Josip Pečarić, Ivan Perić
[3] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, 2nd Edition,
Cambridge University Press, Cambridge, 1967.
vol. 9, iss. 1, art. 9, 2008
[4] M. KRNIĆ AND J. PEČARIĆ, Extension of Hilbert’s inequality, J. Math. Anal.
Appl., 324(1) (2006), 150–160.
Title Page
and Predrag Vuković
[5] M. KRNIĆ AND J. PEČARIĆ, General Hilbert’s and Hardy’s Inequalities, Math.
Inequal. Appl., 8(1) (2005), 29–52.
[6] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London,
1993.
[7] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J.
Math. Anal. Appl., 226 (1998), 166–179.
[8] B. YANG, I. BRNETIĆ, M. KRNIĆ AND J. PEČARIĆ, Generalization of
Hilbert and Hardy-Hilbert integral inequalities, Math. Inequal. Appl., 8(2)
(2005), 259–272.
[9] ZHONGXUE LÜ, Some new inequalities similar to Hilbert-Pachpatte type inequalities, J. Inequal. Pure Appl. Math., 4(2) (2003), Art. 33. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=271].
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